Count Robot Groups solution and explanation
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The problem
You are given a strictly increasing integer array position, where position[i] is the initial position of the ith robot at time t = 0.
You are also given an integer array speed, where speed[i] is the constant speed of the ith robot in units per second, and an integer distance.
Time is continuous and measured in seconds. A robot or group with speed v moves v * t units to the right over any interval of t seconds.
Whenever the distance between two robots or groups becomes at most distance, they merge into a single group.
If multiple robots or groups satisfy the merging condition at the same time, all merges happen simultaneously. In particular, every connected collection of robots or groups whose consecutive positions differ by at most distance merges into one group.
After a merge, the resulting group takes the current position and speed of the rightmost robot in that group. Once merged, robots never separate.
Return the number of groups remaining after all possible merges have occurred.
Example 1:
Input: position = [1,5,6,20], speed = [4,3,2,3], distance = 1
Output: 2
Explanation:
- Initially, the groups are {R1}, {R2}, {R3}, and {R4}.
- At
t = 0, the robots R2 and R3 at positions 5 and 6, respectively, merge because they are 1 unit apart. The resulting group moves with the position and speed of the rightmost robot R3. The groups are now {R1}, {R2, R3}, and {R4}. - Later at
t = 2, the robot R1 catches up to the group {R2, R3} and merges with it. The groups are now {R1, R2, R3} and {R4}.
Thus, the answer is 2.
Example 2:
Input: position = [1,5,9], speed = [3,2,2], distance = 2
Output: 2
Explanation:
- Initially, the groups are {R1}, {R2}, and {R3}.
- At
t = 2, the robot R1 catches up to the robot R2 and merges with it. The resulting group moves with the position and speed of the rightmost robot R2. The groups are now {R1, R2} and {R3}.
Thus, the answer is 2.
Example 3:
Input: position = [9], speed = [8], distance = 5
Output: 1
Explanation:
Initially, there is only one group. Therefore, the answer is 1.
Constraints:
1 <= position.length == speed.length <= 1051 <= position[i], speed[i], distance <= 109positionis strictly increasing.
Approach overview
Problem Overview: You are given the positions of n robots on a line and a distance parameter d. Two robots are in the same group when their distance is at most d. The connection is transitive: if robot A connects to B and B connects to C, all three form a single group. Return the total number of connected robot groups.
Approach 1: Brute Force + Union-Find (O(n²) time, O(n) space)
Build the graph explicitly: for every pair of robots (i, j), compute their distance and call union(i, j) whenever the distance is <= d. After processing all pairs, count the number of distinct roots in the union-find structure. This approach is correct and easy to reason about, but it checks every pair, so it only works for small n. At n = 10^5, that is roughly 5 billion comparisons, which is too slow.
Approach 2: Sort + Sweep (O(n log n) time, O(1) extra space)
Sort the robot positions first. On a sorted line, a gap larger than d between two consecutive robots is an absolute barrier: no robot on the left side can ever reach a robot on the right side. Scan from left to right and start a new group every time positions[i] - positions[i-1] > d. If the gap is <= d, the current robot belongs to the same group as the previous one. This is the optimal approach for the classic one-dimensional connected-components version of the problem. Sorting costs O(n log n), and the single sweep costs O(n). No extra graph or parent array is required.
Approach 3: Sort + Union-Find (O(n log n) time, O(n) space)
If you need actual group IDs or plan to merge robot groups with additional constraints, keep the union-find structure and sort positions. Union only the adjacent pairs whose gap is <= d, then count the distinct roots. The complexity is still dominated by sorting, but the DSU parent array uses O(n) extra space. Use this variant when the follow-up question asks you to return members of each group or maintain extra group metadata.
Recommended for interviews: Use Approach 2. It shows that you understand the one-dimensional property: only consecutive sorted gaps matter. Brute-force union-find proves you can model connected graph components, but it does not scale. The sorted sweep is concise, avoids over-engineering, and gives the expected O(n log n) solution. Better yet, it can also fill group IDs by assigning a counter when a new group starts.
The written solution for this problem is still being prepared. Try it yourself in the meantime.
Try solving it yourselfComplexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force + Union-Find | O(n²) | O(n) | When n is small or you need an easy correctness proof |
| Sort + Sweep | O(n log n) | O(1) | General case for the count-only problem |
| Sort + Union-Find | O(n log n) | O(n) | When you need component IDs or group metadata |
Frequently asked questions
Is Count Robot Groups easy or hard?
It is a Medium difficulty problem. The concept is simple once you realize that two components split only when there is a gap larger than d between consecutive sorted positions. That observation is the main step candidates often miss.
How to solve Count Robot Groups in O(n)?
If the robot positions are already sorted, the sweeping part is O(n). If positions are bounded integers, you can use counting sort or bucket sort to avoid comparison-based sorting, giving an overall O(n) algorithm.
Is Count Robot Groups asked at Google or Amazon?
This exact problem is not a widely reported interview question from a single company. It is frequently used by candidates solving distance-based connected components problems and has obvious interview value at any company that asks sorting or union-find questions.
What data structure is used in Count Robot Groups?
The recommended solution uses an array for sorting and a small integer counter for groups. A union-find data structure is useful when the problem asks you to return group IDs or merge groups with additional data.
What is the best algorithm for Count Robot Groups?
Sort the positions and scan the array once. Whenever the gap between two consecutive positions is greater than d, start a new group. The optimal time complexity is O(n log n) because sorting dominates the solution.
What is the time complexity of Count Robot Groups?
The optimal solution runs in O(n log n) time: O(n log n) for sorting and O(n) for the sweep. The brute-force approach that unions every pair runs in O(n²) time and should not be used for large inputs.
How do I solve Count Robot Groups in Python or Java?
In Python, sort positions with sorted(), then loop through adjacent differences and increment the group count when a gap exceeds d. In Java, use Arrays.sort() and the same linear scan. Both languages need the same O(n log n) solution.